Problem Set 2 is basedon a problem in the MT3Dmanual; also discussed inZ&B, p. 228-231.

2D steady state flow in a confined aquifer

We want to predict thebreakthrough curve at thepumping well. The transportproblem is transient.

Peclet numbers = 5 and 25

Zone of lowhydraulic conductivity

Units in MT3D – see p. 6-8 in the manual

Recommended: use ppm= mg/l gm/m3

That is, use meters; mass is reported in grams.

Mass = c Q t

Cs = 57.87 ppm Cs = 0

0.00

0.20

0.40

0.60

0.80

1.00

1.20

0.00 0.20 0.40 0.60 0.80 1.00 1.20

Time (years)

Co

nce

ntr

atio

n

TVD

HMOC

Upstream weighting

Central FD

Upstream FD

Central FD

TVD

NOTE. These results were produced using an old version of MT3DMS.Please run again with the latest version of the code.

MT3DMS Solution Options

1

2

3

4

PS#2

Central Difference Solution

Time step multiplier = 141 time steps

Time step multiplier = 1.213 time steps

See information on solution methodologies underthe MT3DMS tab on the course homepage for moreabout these parameters.

Courant number

Boundary Conditions

Need to designatethese boundarycells as inactiveconcentration cells.Use zone 10 inthe diffusionsproperties menu ofGroundwater Vistas.

Solution at t=1 year

Mass Balance Considerations in MT3DMS

Sources of mass balance information:*.out file*.mas filemass balance summary in GW Vistas

See supplemental information for PS#2 postedon the course homepage for more information onmass balance options.

Mass Balance states that:Mass IN = Mass OUTwhere changes in mass storageare considered either as contributionsto mass IN or to mass OUT.

Water Flow:IN= through upper boundary; injection wellOUT= pumping well; lower boundary

Mass Flux:IN= through injection well; changes in storageOUT= pumping well; lower boundary; changes in storage

wells

IN - OUT = S where S = 0 at steady state conditions

From the *.out file (TVD solution)

Mass Storage: Water

Consider a cell in the model

IN - OUT = S where change in storage isS = S(t2) – S(t1)

If IN > OUT, the water level rises andthere is an increase in mass of water in the cell.IN = OUT + S, where S is positive.Note that S is on the OUT side of the equation.

If OUT > ININ – S = OUT, where S is negativeS is on the IN side of the equation.

From the *.out file (TVD solution)

S

S = c (x y z )

Mass Storage: Solute

IN - OUT = S where change in storage isS = S(t2) – S(t1)

If IN > OUT, concentration in cell increases andthere is an increase in solute mass in the cell.IN = OUT + S, where S is positive.Note that S is on the OUT side of the equation.There is an apparent “sink” inside the cell.

If OUT > IN, the concentration in cell decreases andthere is a decrease in solute mass in the cell.IN – S = OUT, where S is negative and S is on the IN side of the equation. There is anapparent “source” inside the cell.

From the *.out file (TVD solution)

S

IN – OUT = 0(INsource+SIN) - (OUTsource + SOUT)= 0

SIN - SOUT = Storage

HMOC *.mas file

q’s =

MOC methods typically report high mass balance errors,especially at early times.

TVD Solution

From the *.out file (TVD solution)

S

IN – OUT = 0(INsource+SIN) - (OUTsource + SOUT)= 0

SIN - SOUT = Storage

From the *.out file (TVD solution)

Last t = 0.0089422 yr

Mass Flux = (mass at t2 - mass at t1) / t